Define a function:
\begin{align} M\left({m}\right): \mathbb{R} &\to \mathbb{C} \\ t &\mapsto \lim_{k \to \infty} \sum_{s \in \{s : m(s) \ne 0, s \le t\}}m(s) k e^{k(s - t)}. \label{eqn:pointmass} \end{align}
$m$ is a function from times t to values, and $M$ maps $m$ to another function from times $t$ to values.
I hope that this definition of $M$ makes it so that for any $t_1 < t_2$, \begin{align} \int_{t_1}^{t_2}\left[M(m)\right](t) \,dt = \sum_{t \in \{t : m(t) \ne 0, t_1 \le t < t_2\}}m(t). \end{align}
Sometimes, I end up doing some manipulations, and I end up with \begin{align} \int_{t_1}^{t_2} \frac{dM(m)}{dt} \cdot x\,dt \end{align} where $x$ is also a function of $t$.
Am I supposed to replace this with $-\sum_{s \in \{s : m(s) \ne 0, t_1 \le s < t_2\}}\frac{dx}{dt} m(s)$? Or zero?
Let us try an example. Correct me if I am making a wrong interpretation.
Say we have $m$ defined as $m(1) = 2, m(3) = 7$ and $m(s) = 0$ for all other $s$. Then
For $t<1$ we have $\{s : m(s) \ne 0, s \le t\}$ is empty, so $M(m)(t) = 0$.
For $t=1$ we have $\{s : m(s) \ne 0, s \le t\} = \{1\}$, so $$ M(m)(1) = \lim_{k\to\infty} 2 k e^{k 0} = \infty $$ For $1 < t < 3$ we have $\{s : m(s) \ne 0, s \le t\} = \{1\}$, so $$ M(m)(1) = \lim_{k\to\infty} 2 k e^{k (1-t)} = 0 $$ For $t = 3$ we have $\{s : m(s) \ne 0, s \le t\} = \{1,3\}$, so $$ M(m)(3) = \lim_{k\to\infty} \big(2 k e^{k (1-3)} + 7 k e^{k 0}\big) = \infty $$ For $t > 3$ we have $\{s : m(s) \ne 0, s \le t\} = \{1,3\}$, so $$ M(m)(t) = \lim_{k\to\infty} \big(2 k e^{k (1-t)} + 7 k e^{k (3-t)}\big) = 0 $$
As written, then, we have all integrals $$ \int_{t_1}^{t_2}\left[M(m)\right](t) \,dt = 0 $$ since the integrand is zero almost everywhere. So what you wrote is not what you want.
It seems, perhaps, you want the limit outside the integral: $$ \lim_{k \to \infty} \int_{t_1}^{t_2}\sum_{s \in \{s : m(s) \ne 0, s \le t\}}m(s) k e^{k(s - t)}\;dt = \sum_{t \in \{t : m(t) \ne 0, t_1 \le t < t_2\}}m(t) $$